Simplify and expand the following expression: $ \dfrac{2t + 1}{t + 8}-\dfrac{t}{5t - 2} $
In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(t + 8)(5t - 2)$ Multiply the first term by $\dfrac{5t - 2}{5t - 2}$ $ \begin{align*} \dfrac{2t + 1}{t + 8} \times \dfrac{5t - 2}{5t - 2} & = \dfrac{(2t + 1)(5t - 2)}{(t + 8)(5t - 2)} \\ & = \dfrac{10t^2 + t - 2}{(t + 8)(5t - 2)}\end{align*} $ Multiply the second term by $\dfrac{t + 8}{t + 8}$ $ \begin{align*} \dfrac{t}{5t - 2} \times \dfrac{t + 8}{t + 8} & = \dfrac{(t)(t + 8)}{(5t - 2)(t + 8)} \\ & = \dfrac{t^2 + 8t}{(5t - 2)(t + 8)}\end{align*} $ Now we have: $ = \dfrac{10t^2 + t - 2}{(t + 8)(5t - 2)} - \dfrac{t^2 + 8t}{(5t - 2)(t + 8)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{10t^2 + t - 2 - (t^2 + 8t)}{(t + 8)(5t - 2)} $ $ = \dfrac{10t^2 + t - 2 - t^2 - 8t}{(t + 8)(5t - 2)} $ $ = \dfrac{9t^2 - 7t - 2}{(t + 8)(5t - 2)}$ Expand the denominator: $ = \dfrac{9t^2 - 7t - 2}{5t^2 + 38t - 16}$